Half Angle Formulas/Cosine

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Theorem

\(\displaystyle \cos \frac \theta 2\) \(=\) \(\displaystyle + \sqrt {\frac {1 + \cos \theta} 2}\) for $\dfrac \theta 2$ in quadrant I or quadrant IV
\(\displaystyle \cos \frac \theta 2\) \(=\) \(\displaystyle - \sqrt {\frac {1 + \cos \theta} 2}\) for $\dfrac \theta 2$ in quadrant II or quadrant III

where $\cos$ denotes cosine.


Proof 1

\(\displaystyle \cos \theta\) \(=\) \(\displaystyle 2 \cos^2 \frac \theta 2 - 1\) Double Angle Formula for Cosine: Corollary 1
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 \cos^2 \frac \theta 2\) \(=\) \(\displaystyle 1 + \cos \theta\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \cos \frac \theta 2\) \(=\) \(\displaystyle \pm \sqrt {\frac {1 + \cos \theta} 2}\)


We also have that:

In quadrant I, and quadrant IV, $\cos \dfrac \theta 2 > 0$
In quadrant II and quadrant III, $\cos \dfrac \theta 2 < 0$.

$\blacksquare$


Proof 2

Define:

$u = \dfrac \theta 2$


Then:

\(\displaystyle \cos^2 u\) \(=\) \(\displaystyle \frac {1 + \cos2u} 2\) Power Reduction Formulas
\(\displaystyle \implies \ \ \) \(\displaystyle \cos \frac \theta 2\) \(=\) \(\displaystyle \pm \sqrt {\frac {1 + \cos \theta} 2}\)


We also have that:

In quadrant I, and quadrant IV, $\cos \dfrac \theta 2 > 0$
In quadrant II and quadrant III, $\cos \dfrac \theta 2 < 0$.

$\blacksquare$


Also see


Sources